UNIT1: COMPUTERNUMBER SYSTEM

Transcription

1 UNIT1: COMPUTERNUMBER SYSTEM Binary,Decimal, Octal and Hexadecimal Number-Base Conversions Addition, Subtraction and Multiplication of Computer Number System Compliment (1 s, 2's and r s complement) Signed Binary Number Binary Arithmetic Page 1 of 35

2 COMPUTER NUMBER SYSTEM 1.1Binary,Decimal, Octal and Hexadecimal 1. Introduction of Computer Number System: We used the man in the long history of different numerical systems and our system we call the decimal system. Numerical system 1. Binary (0,1) 2 2. Decimal (0,to 91) Octal(0 to 7) 8 4. Hexadecimal(0 to 9 ABCDEF) 2 النظام الثناي ي : System Binary This means that the code contains only the digits 0 and 1 and uses this code to convert any data to write the number 1 and 0. النظام العشري Decimal System The decimal system, if the basis of (10) and the following figures by category (0,1,2,3,4,5,6,7,8,9) النظام الثماني Octal System Octal number system is a counting system with a basis of the number 8, and uses the numbers from 0 to 7. سداسي عشريSystem Hexadecimal A Hexadecimal Number is based on the number 16 As well as the familiar digits 0 to 9, there are also the letters "A","B","C","D","E" and "F" in place of the decimal numbers 10 to 15. Page 2 of 35

3 1.2Number-Base Conversions Q1. Convert Binary to Decimal Number Systems? ( ) 2 = ( ) 10 Ans: x 16 8 x = 87 ( ) 2 = (87) 10 Q2. Convert Decimal to Binary Number System (274) 10 = ( ) 2 Ans: x 64 x 32 x 16 8 x 4 x 2 1 x = 274 (274) 10 = ( ) 2 Q3. Convert Binary to Octal Number Systems? ( ) 2 = ( ) 8 Ans: Add leading zeros ( ) 2 = (146) 8 Page 3 of 35

4 Q4 Convert Octal to Binary Number Systems? (6132) 8 = ( ) 2 Ans: إجابة 6 = = = = 010 ( ) 2 (6132) 8 = ( ) 2 Q5. Convert Binary to Hexadecimal number systems? ( ) 2 =(------) 16 Ans: ( ) 2 =(56) 16 Q6 Convert Hexadecimal to Binary number systems? (39F) 16 = ( ) 2. Ans: 3 = = 1001 F = 1111 (39F) 16 = ( ) 2 Page 4 of 35

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6 ثناي یالحسابArithmetic 2. Binary Binary arithmetic is essential part of all the digital computers and many other digital system. الثناي یةزاي د 2.1 Binary Addition It is a key for binary subtraction, multiplication, division. There four rules of the binary addition. In fourth case, a binary addition is creating a sum of (1+1=10) i.e. 0 is write in the given column and a carry of 1 over to the next column. EXAMPLE - ADDITION ناقص الثناي یةSubtraction 2.2Binary Subtraction and Borrow, these two words will be used very frequently for the binary subtraction.there four rules of the binary substration.there four rules of the binary Subtraction. Page 6 of 35

7 EXAMPLE - SUBTRACTION الثناي یةضربMultiplication 2.3 Binary Binary multiplication is similar to decimal multiplication. It is simpler than decimal multiplication because only 0s and 1s are involved.there four rules of the binary multiplication. EXAMPLE - MULTIPLICATION Page 7 of 35

8 2.4 Binary Division Binary division is similar to decimal division. It is called as the long division procedure. EXAMPLE - DIVISION 3. Octal Arithmetic Octal Number System Following are the characteristics of an octal number system. Uses eight digits, 0,1,2,3,4,5,6,7. Also called base 8 number system Each position in a octal number represents a 0 power of the base (8). Example 8 0 Last position in a octal number represents a x power of the base (8). Example 8 x where x represents the last position - 1. Page 8 of 35

9 EXAMPLE Octal Number: Calculating Decimal Equivalent: Step Octal Number Decimal Number Step ((1 x 8 4 ) + (2 x 8 3 ) + (5 x 8 2 ) + (7 x 8 1 ) + (0 x 8 0 )) 10 Step ( ) 10 Step Note: is normally written as Octal Addition Page 9 of 35

10 3.2 Octal Subtraction 3.2 Octal Multiplication Page 10 of 35

11 4. Hexadecimal Arithmetic Hexadecimal Number System Following are the characteristics of a hexadecimal number system. Uses 10 digits and 6 letters, 0,1,2,3,4,5,6,7,8,9,A,B,C,D,E,F. Letters represents numbers starting from 10. A = 10. B = 11, C = 12, D = 13, E = 14, F = 15. Also called base 16 number system Each position in a hexadecimal number represents a 0 power of the base (16). Example 16 0 Last position in a hexadecimal number represents a x power of the base (16). Example 16 x where x represents the last position - 1. EXAMPLE Hexadecimal Number: 19FDE 16 Calculating Decimal Equivalent: Step Hexadecimal Number Decimal Number Step 1 19FDE 16 ((1 x 16 4 ) + (9 x 16 3 ) + (F x 16 2 ) + (D x 16 1 ) + (E x 16 0 )) 10 Step 2 19FDE 16 ((1 x 16 4 ) + (9 x 16 3 ) + (15 x 16 2 ) + (13 x 16 1 ) + (14 x 16 0 )) 10 Step 3 19FDE 16 ( ) 10 Step 4 19FDE Note: 19FDE 16 is normally written as 19FDE. Page 11 of 35

12 4.1 Hexadecimal Addition Page 12 of 35

13 4.2 Hexadecimal Subtraction and Multiplication Page 13 of 35

14 تكملالحسابArithmetic 1.3 Complement Complements are used in the digital computers in order to simplify the subtraction operation and for the logical manipulations. For each radix-r system (radix r represent base of number system) there are two types of complements S.N. Complement 1 Radix Complement Description The radix complement is referred to as the r's complement 1 Diminished Radix Complement The diminished radix complement is referred to as the (r-1)'s complement Binary system complements As the binary system has base r = 2. So the two types of complements for the binary system are 2's complement and 1's complement. 1's complement The 1's complement of a number is found by changing all 1's to 0's and all 0's to 1's. This is called as taking complement or 1's complement. Example of 1's Complement is as follows. Page 14 of 35

15 2's complement The 2's complement of binary number is obtained by adding 1 to the Least Significant Bit (LSB) of 1's complement of the number. 2's complement = 1's complement + 1 Example of 2's Complement is as follows. Page 15 of 35

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22 R s AND (R-1) s COMPLEMENT OF NUMBER SYSTEMS In digital system complement is used to find subtraction of number base system and for digital manipulation. If R be the base of a number system then that number system can have two complements respectively R s and (R-1) s complement. The base of binary number system is 2 so there can be 2 s complement and 1 s complement of this system. Similarly in hexadecimal number system there are 15 s complement and 16 s complement. Now the general formula for find R s complement and (R-1) s are given below: R s complement of any number system = (R n ) 10 -N and (R-1) s complement of any number system = {(R n ) 10-1)}-N Where R = Radix or base of that number system n = Number of digits in the number and N = The given number R s AND (R-1) s COMPLEMENT OF BINARY NUMBER SYSTEM The base of binary number system is 2 so for binary number R=2 and R-1 = 1 thus 1 s complement for binary numbers is {(2 n ) 10-1)}-N and 2 s complement is (2 n ) 10 -N. Example: Find 1 s complement and 2 s complement of the binary number ? Solution: 1 s complement of = {(2 7 ) 10-1)} = = = s complement of = (2 7 ) 10 -N = = = Page 22 of 35

23 There is a short cut method for finding 1 s complement and 2 s complement of binary numbers. We just need to alter 0 s by 1 s and 1 s by 0 s for 1 s complement and for 2 s complement find 1 s complement from this method then add 1 to LSB(least significant bit). So the above problem can be solved as: 1 s complement of = s complement of = 1 s complement of the binary number + 1 = = R s AND (R-1) s COMPLEMENT OF BASE 3 NUMBER SYSTEM Here the base is 3 so R = 3 and R-1 = 2 thus 3 s complement of base 3 number system = (3 n ) 10 -N and 2 s complement of base 3 number system = {(3 n ) 10-1)}-N. Example: Find 3 s complement and 2 s complement of 121 3? Solution: 3 s complement and 2 s complement is found as: 2 s complement of = {(3 3 ) 10-1)} = = = s complement of = (3 n ) 10 -N = = = There is a shortcut method for finding 2 s and 3 s complement of this number base system. Subtract each digit of the given number from 2 and take them together. Thus the value obtained is the 2 s complement of the given number. And for 3 s complement just add 1 to the LSB. Example: Find 3 s complement and 2 s complement of 121 3? Solution: 2 s complement of = Page 23 of 35

24 = And 3 s complement of = 2 s complement + 1 = = R s AND (R-1) s COMPLEMENT OF BASE 4 NUMBER SYSTEM In this system base is 4 so R = 4 and R-1 = 3 thus 4 s complement in this number system is (4 n ) 10 -N and 3 s complement in this system is {(4 n ) 10-1)}-N. Example: Find the 3 s complement and 4 s complement of 130 4? Solution: 3 s complement of = {(4 3 ) 10-1)} = = = = s complement of = (4 3 ) = = = = The shortcut method for this system is as follows: first subtract each bit of number from 3 and then take them together for 3 s complement and for 4 s complement add 1 to the LSB. Example: Find the 3 s complement and 4 s complement of 130 4? Solution: 3 s complement of = = s complement of = 3 s complement of Page 24 of 35

25 = = R s AND (R-1) s COMPLEMENT OF BASE 5 NUMBER SYSTEM This number system has 4 s and 5 s complement so 4 s complement for this number system is {(5 n ) 10-1)}-N and 5 s complement is (5 n ) 10 -N. Example: Find the 4 s complement and 5 s complement of 224 5? Solution: 4 s complement of = {(5 3 ) 10-1)} = = = = s complement of = (5 3 ) = = = =221 5 The shortcut method for 4 s complement is as: we have to subtract each digit of the given number from 4. And for 5 s complement just add 1 to the 4 s complement. 4 s complement of = = s complement of = 4 s complement of = = R s AND (R-1) s COMPLEMENT OF BASE 6 NUMBER SYSTEM 6 is the base of this system so R = 6 and R-1 = 5 thus 6 s complement is equal to Page 25 of 35

26 (6 n ) 10 -N and 5 s complement is equal to {(6 n ) 10-1)}-N. Example: Find 5 s complement and 6 s complement of (530) 6? Solution: 5 s and 6 s complement of is found as: 5 s complement of = {(6 3 ) 10-1)} = = = = s complement of = (6 3 ) = = = = 26 6 By shortcut method we can solve it as we have solved the above problems of base 3, 4 or 5. In this system the starting digit is 0 and the last digit is 5 so repeat the above method. 5 s complement of = = s complement of = 5 s complement of = = 26 6 R s AND (R-1) s COMPLEMENT OF BASE 7 NUMBER SYSTEM The base 7 number system has 6 s and 7 s complement so R = 7 and R-1 = 6 thus 6 s complement is {(7 n ) 10-1)} N and 7 s complement is (7 n ) 10 N. Example: Find the 6 s complement and 7 s complement of (653) 7? Solution: 6 s complement of = {(7 3 ) 10-1)} Page 26 of 35

27 = = = = s complement of = (7 3 ) = = = = 14 7 The above problem can also be solved by a shortcut method as: subtract each digit of the given number from 6 and then take them as a whole number. For 7 s complement add one to the 6 s complement of the given number. Thus the above problem is solved as: 6 s complement of = = s complement of = 6 s complement of = = 14 7 R s AND (R-1) s COMPLEMENT OF OCTAL NUMBER SYSTEM In octal number system there is 7 s complement and 8 s complement because here R = 8 and R-1 = 7. Now 7 s complement of octal number is {(8 n ) 10-1)} N and 8 s complement of octal number is (8 n ) 10 N. Example: Find 7 s complement and 8 s complement of the octal number (172) 8? Solution: The 7 s and 8 s complement of is found as: 7 s complement of = {(8 3 ) 10-1)} = Page 27 of 35

28 = = = s complement of = (8 3 ) = = = = In short the above problem can be solved by subtracting each digit from 7 for 7 s complement and add 1 to the 7 s complement to find 8 s complement of given octal number. 7 s complement of = = s complement of = 7 s complement of = = R s AND (R-1) s COMPLEMENT OF BASE 9 NUMBER SYSTEM The base 9 number system has R = 9 and R-1 = 8 so 8 s complement and 9 s complement of base 9 number system are {(9 n ) 10-1)} N and (9 n ) 10 N respectively. Example: Find 8 s complement and 9 s complement of 807 9? Solution: 8 s complement of = {(9 3 ) 10-1)} = = = = 81 9 Page 28 of 35

29 9 s complement of = (9 3 ) = = = = 82 9 The shortcut method for this problem is shown below: 8 s complement of = = s complement of = 8 s complement of = = 82 9 R s AND (R-1) s COMPLEMENT OF DECIMAL NUMBER SYSTEM The base of decimal number system is 10 since there are 10 distinguish digits in this system starting from 0 to 9. In this system R = 10 and R-1 = 9 thus 9 s complement of decimal number is {(10 n ) 10-1)} N and 10 s complement is (10 n ) 10 N. Example: Find 9 s and 10 s complement of the decimal number ? 9 s complement = {(10 3 ) 10-1)} = = s complement = (10 3 ) = = The shortcut method for 9 s complement and 10 s complement of decimal number is shown below: 9 s complement Page 29 of 35

30 = = s complement = 9 s complement = = R s AND (R-1) s COMPLEMENT OF BASE 11 NUMBER SYSTEM In this system there are 11 digits. The last digit of this system is represented by the English alphabet A which is equivalent to 10 in decimal number system. Here R = 11 and R-1 = 10. Thus 10 s complement of this number system is {(11 n ) 10-1)} N and 11 s complement of this system is (11 n ) 10 N. Example: Find 10 s complement and 11 s complement of 1A1 11? Solution: 10 s complement and 11 s complement of 1A1 11 is found as: 10 s complement of 1A1 11 = {(11 3 ) 10-1)} 1A1 11 = A1 11 = = = s complement of 1A1 11 = (11 3 ) 10 1A1 11 = A1 11 = = = 90A 11 To find 10 s complement of 1A1 11 by shortcut method we subtract each digit of the given number from A which is equal to 10. And for 11 s complement add 1 to the value of 10 s complement of 1A1 11. So the given problem is solved as: 10 s complement of 1A1 11 = AAA 1A1 = s complement of 1A1 11 = 10 s complement of 1A Page 30 of 35

31 = = 90A 11 R s AND (R-1) s COMPLEMENT OF BASE 12 NUMBER SYSTEM Here the base of the number system is 12. In this system the last two digits are represented by A and B which are equivalent to 10 and 11. The digits are as: 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, A and B. Here R = 12 and R-1 = 11 thus 11 s complement is equal to {(12 n ) 10-1)} N and 12 s complement is equal to (12 n ) 10 N. Example: Find 11 s complement and 12 s complement of AB0 12? Solution: 11 s complement and 12 s complement of AB0 12 is as follows: 11 s complement of AB0 12 = {(12 3 ) 10-1)} AB0 12 = AB0 12 = = = 10B s complement of AB0 12 = (12 3 ) 10 AB0 12 = AB0 12 = = = We can solve these problems by another method: for 11 s complement subtract each digit from B and for 12 s complement add 1 to 11 s complement of the given number. 11 s complement of AB0 12 = BBB AB0 = 10B s complement of AB0 12 = 11 s complement of AB = 10B = Page 31 of 35

32 R s AND (R-1) s COMPLEMENT OF BASE 13 NUMBER SYSTEM In this system R = 13 and R-1 = 12 so 12 s complement and 13 s complements are as follows: 12 s complement of base 13 number system = {(13 n ) 10-1)} N and 13 s complement of base 13 number system = (13 n ) 10 N. Example: Find 12 s complement and 13 s complement of ABC 13? Solution: 12 s complement of ABC 13 = {(13 3 ) 10-1)} ABC 13 = ABC 13 = = = s complement of ABC 13 = (13 3 ) 10 ABC 13 = = ABC 13 = = = By shortcut method: 12 s complement of ABC 13 = CCC ABC = s complement of ABC 13 = 12 s complement of ABC = = R s AND (R-1) s COMPLEMENT OF BASE 14 NUMBER SYSTEM For base 14 number system R = 14 and R-1 = 13 so 14 s complement of base 14 number system is (14 n ) 10 N and 13 s complement of this system is {(14 n ) 10-1)} N. Example: Find 14 s complement and 13 s complement of D15 14? Page 32 of 35

33 Solution: Here base is 14 so let s use above two formulas to find 14 s and 13 s complement of D s complement of D15 14 = (14 3 ) 10 D15 14 = D15 14 = = = C s complement of D15 14 = {(14 3 ) 10-1)} D15 14 = D15 14 = = = C8 14 Shortcut method for this problem is as follows: (D = 13) 13 s complement of D15 14 = DDD D15 = C s complement of D15 14 = 13 s complement of D = C = C9 14 R s AND (R-1) s COMPLEMENT OF BASE 15 NUMBER SYSTEM Base 15 number system has 15 s complement and 14 s complement which are as follows: 15 s complement of base 15 = (15 n ) 10 N 14 s complement of base 15 = {(15 n ) 10-1)} N Example: Find 15 complement and 14 s complement of AE0 15? Solution: The solution of the above problem using the above two formulas: Page 33 of 35

34 14 s complement of AE0 15 = {(15 3 ) 10-1)} AE0 15 = AE0 15 = = = 40E s complement of AE0 15 = (15 3 ) 10 AE0 15 = AE0 15 = = = Shortcut method: 14 s complement of AE0 15 = EEE AE0 = 40E s complement of AE0 15 = 14 s complement of AE = 40E = R s AND (R-1) s COMPLEMENT OF HEXADECIMAL NUMBER SYSTEM Hexadecimal number system has 15 s complement and 16 s complement. Here R = 16 and R-1 = 15 thus we can write: 15 s complement of hexadecimal number = {(16 n ) 10-1)} N 16 s complement of hexadecimal number = (16 n ) 10 N Example: Find 15 s complement and 16 s complement of the hexadecimal number F9 16? Page 34 of 35

35 Solution: 15 s complement of hexadecimal number F9 16 = {(16 2 ) 10-1)} F9 16 = F9 16 = = 6 10 = s complement of hexadecimal number F9 16 = (16 2 ) 10 F9 16 = F9 16 = = 7 10 = 7 16 Shortcut method: 15 s complement of hexadecimal number F9 16 = FF F9 = s complement of hexadecimal number F9 16 = 15 s complement of hexadecimal number F = = 7 16 Page 35 of 35